Symmetric functions and spherical t-designs in $\pmb {\mathbb {R}^2}$

Abstract

Given k points in $S^1$ satisfying certain conditions which are determined through their symmetric functions, we introduce a method for constructing spherical t-designs in $\mathbb {R}^2$ with $2t+k$ elements. This approach points toward a better understanding of the space of spherical t-designs as well as provides a systematic way to obtain examples.

1 Introduction

A finite subset X of $S^{n-1}$ in $\mathbb {R}^n$ is a spherical t-design if for any polynomial $f(\mathbf{x})=f(x_1,\ldots ,x_n)$ of degree at most t, the value of the integral of $f(\mathbf{x})$ on $S^{n-1}$ divided by the volume of $S^{n-1}$ equals the average value of $f(\mathbf{x})$ on the finite set X. They were introduced by Delsarte et al. [5]. In the survey [1] a detailed explanation of the developments in the subject can be found, along with a discussion of the connections of spherical t-designs with other fields of mathematics such as group theory, number theory and orthogonal polynomials, among others.

The case of spherical t-designs in $\mathbb {R}^2$ was studied in [8]. Consider the elements of $\mathbb {R}^2$ as complex numbers, thus

$$\begin{aligned} S^1=\{z\in \mathbb {C}: |z|=1\}. \end{aligned}$$

Denote $\sigma _i$, $i=1,\ldots ,k$ the elementary symmetric polynomials (the symmetric functions, for short) of a set of k elements $\{z_1,\ldots ,z_k\}$. We usually write $\sigma _i$ instead of $\sigma _i(z_1,\ldots ,z_k)$. A spherical t-design X in $\mathbb {R}^2$ can be defined by means of the complex polynomial having the elements of X as its roots (for this and other equivalent definitions of spherical t-designs, see [1]).

Definition 1

Let t, n be positive integers, and $n\ge t+1$. A set $X=\{z_1,\ldots ,z_n\}\subset S^1$ is a spherical t-design, or simply a t-design, if

$$\begin{aligned} \sigma _i(z_1,\ldots ,z_n)=0 \text { for } i=1,\ldots ,t. \end{aligned}$$

(1)

Equivalently, the polynomial

$$\begin{aligned} f(z)=(z-z_1)\cdots (z-z_n)=z^n+c_1z^{n-1}+\cdots +c_{n-1}z+c_n \end{aligned}$$

(2)

is such that $c_i=0$ for $i=1,\ldots ,t$, because $c_i=(-1)^i\sigma _i$.

As the elements of a given t-design X are unit-norm complex numbers, we can observe the following:

Remark 2

(cf. [8, Lemma 2])

• (i) The condition $\sigma _i=0$ for $i=1,\ldots ,t$ is equivalent to $\sigma _{n-i}=0$ for $i=1,\ldots ,t$.

• (ii) $\overline{\sigma _i}\sigma _n=\sigma _{n-i}$, for $i=1\ldots ,n$ ($\sigma _0$ is taken as 1).

Notice that this definition of t-design allows repeated elements in the set X.

It is clear that the n-th roots of any unit-norm complex number (regular n-gons) form a t-design if $n>t$. Moreover, it is easy to see that, if $n_1,\ldots ,n_s$ are positive integer numbers greater than t and $w_1,\ldots ,w_s\in S^1$, then the roots of the polynomial

$$\begin{aligned} f(z)=\prod _{i=1}^s(z^{n_i}-w_i), \end{aligned}$$

(3)

also form a t-design with $n=\sum _{i=1}^s n_i$ elements. In [8], spherical t-designs as such are called group-type t-designs, while a t-design that is not a group-type t-design is called a non-group-type t-design. In particular, for any positive integer t there exists always group-type t-designs with n elements, provided $n\ge t+1$. The natural question of deciding whether there are t-designs besides these easily obtained ones is answered by the following theorem.

Theorem 3

[8, Theorem A] Let X be a spherical t-design in $\mathbb {R}^2$ with $|X|=n$. Then

• (i) for $t+1\le n\le 2t+2$, X is always a group-type t-design; more precisely, when $t+1\le n\le 2t+1$, X is a regular n-gon and when $n=2t+2$, X is the union of two regular $(t+1)$-gons (this includes the $(2t+2)$-gon case);

• (ii) for each $n\ge 2t+3$, besides group-type t-designs, there are as many as $\aleph _1$ non-group-type t-designs.

This paper goes deeper in the understanding of t-designs in $\mathbb {R}^2$ as started in [8], though our approach is different. We can say that, in some way, we parameterize the design with some of its points. The precise information to make this possible turn out to be in the symmetric functions of this points, more specifically, on algebraic and metric conditions upon such functions. Moreover, group-type t-designs can be obtained through easily described algebraic conditions.

In Sect. 4 our general method to construct a t-design X with $|X|=2t+k$, given $a_1,\ldots ,a_k\in S^1$ is given. The $a_i's$ will belong to X while the other 2t elements in the design turn out to be the roots of a polynomial whose coefficients are expressed in terms of the symmetric functions $\sigma _1,\ldots ,\sigma _k$ of the $a_i's$. Such polynomial will be denoted by $G_{k,t}$. So, conditions must be imposed on $a_1,\ldots ,a_k$ in order that: (1) the coefficients of the polynomial $G_{k,t}$ are well defined and, (2) the roots of $G_{k,t}$ are all unit-norm complex numbers. When these conditions are satisfied we will say that the $a_i's$ are in good position. Using this method, we construct a family of 3-designs in Example 34.

For clarity in the exposition, our method is illustrated first in Sect. 3 by means of the simplest case where a non-group spherical t-design can appear: a spherical 2-design of 7 elements. In Sect. 3.1 we show how conditions on three unit-norm complex numbers naturally appear when trying to construct a spherical 2-design of 7 elements containing them. Proposition 14 assures the existence and unicity of such a design under the obtained conditions and Example 18 gives a family of non-group-type 2-designs. In this case we are able to give a complete description of our object of study: Proposition 22 states that any 2-design with 7 elements, where at least five of them are distinct, can be obtained by this method. In Sect. 3.3 conditions for a 2-design with 7 elements to be of group-type are given and this result is generalized latter for any t and $k=t+1$, in Proposition 37.

Another reason for giving a separate treatment to 2-designs is that they are an important kind of unit norm tight frames (see [2]), also known as balanced unit norm tight frames, which are studied in [7].

Our results indicate that it makes sense to consider the space of t-designs in the $\sigma$-space, that is, the space whose coordinates are the symmetric functions. This is explained in Sect. 5 and applied to the case of 2-designs. Further development of this approach could contribute to a new understanding of spherical t-designs.

2 Polynomials with roots on the unit circle

In this article, we are faced with the problem of deciding whether a given complex polynomial has all of its roots in $S^1$. This is not a trivial issue and it is particularly difficult to find necessary and sufficient conditions on the coefficients of the polynomials (see, for instance, [10] and [3]).

Definition 4

A polynomial $P(z)=\sum _{k=0}^m \alpha _kz^k\in \mathbb {C}[z]$ is self-inversive if $\alpha _{m-k}=w\overline{\alpha _k}$ for $k=0,\ldots ,m$, where $w\in \mathbb {C}$, $|w|=1$.

It is easily seen that a polynomial with all of its roots in the unit circle is self-inversive, but the following classical result says much more:

Theorem 5

(Cohn [3]) A complex polynomial P has all of its zeros in the unit circle if and only if the following two conditions holds:

• i) P is self-inversive,

• ii) All zeros of $P'$ lie in the unit disk.

It is also not an easy task to decide whether a given polynomial has its roots on the unit disk. However, for the case of degree 3, we can give the following criteria.

Proposition 6

Let $g(z)=z^3+az^2+bz+c\in \mathbb {C}[z]$ such that $c-ba\ne 0$ and let $D=\{z\in \mathbb {C}: |z|\le 1\}$. If $|a|<1$ and $|c-ab|<1-|a|^2-\left| \overline{a}c-b\right|$, then g(z) has all of its roots in D.

If $c=ba$ then the roots of g(z) lie in D if and only if $a,b\in D$.

Proof

Observe that z is a root of g(z) if and only if

$$\begin{aligned} z^2=\frac{-bz-c}{z+a}. \end{aligned}$$

By hypothesis, the mapping $M:\mathbb {C}_{\infty }\rightarrow \mathbb {C}_{\infty }$ given by $M(z)=\dfrac{-bz-c}{z+a}$ is a Möbius transformation (see, for instance, [4]) such that, if $|a|\ne 1$, $M(S^1)$ is a circle C with center $z_0=-\dfrac{\overline{a}c-b}{|a|^2-1}$ and radius $\rho =\dfrac{|c-ab|}{||a|^2-1|}$.

If $|a|<1$ the outside of D is mapped to the inside of C, while the second hypothesis is equivalent to $\rho <1-|z_0|$, which implies that C and its interior are contained in D. So, if z is a root of g(z) and $|z|>1$, then $M(z)=z^2\in D$, a contradiction.

The last affirmation is obvious since $z^3+az^2+bz+ab=(z^2+b)(z+a)$. $\square$

The following result, due to Lakatos and Losonczi [9] (see also [10]) gives sufficient conditions for a self-inversive polynomial to have all of its roots on the unit circle.

Theorem 7

(Lakatos–Losonczi) All zeros of a self-inversive polynomial $P(z)=\sum _{k=0}^m \alpha _kz^k\in \mathbb {C}[z]$ of degree $m\ge 1$ are on the unit circle if

$$\begin{aligned} |\alpha _m|\ge \frac{1}{2}\sum _{k=1}^{m-1}|\alpha _k|. \end{aligned}$$

(4)

For real polynomials with roots on the unit circle, the relation between them and their coefficients can be clarified: Let $b=(b_1,\ldots ,b_d)\in \mathbb {R}^d$ and denote $g_b$ the polynomial

$$\begin{aligned} g_b=(x^{2d}+1)+b_1(x^{2d-1}+x)+\cdots +b_{d-1}(x^{d+1}+x^{d-1})+b_dx^d. \end{aligned}$$

It is not difficult to see that if $g\in \mathbb {R}[z]$ is a monic polynomial of degree 2d such that all of its complex roots lie on $S^1$ and 1 and $-1$ are roots of g with even multiplicity (possibly 0), then $g=g_b$ for some $b\in \mathbb {R}^d$. Moreover, let $V_d$ be the set of those $b\in \mathbb {R}^d$ such that all of the roots of $g_b$ lie in $S^1$ and

$$\begin{aligned} I_d=\{(r_1,\ldots ,r_d)\in \mathbb {R}^d:-2\le r_1\le r_2 \le \cdots \le r_d\le 2\}. \end{aligned}$$

The following proposition is Lemma 2.1.1 in [6].

Proposition 8

$V_d$ is homeomorphic to $I_d$.

Remark 9

The homemorphism $\varPhi$ of Proposition 8 is given in the following way: for $r=(r_1,\ldots ,r_d)\in I_d$, define $\varPhi (r)$ to be the unique $b\in \mathbb {R}^d$ such that

$$\begin{aligned} x^d \left( \sum _{i=0}^d(-1)^i\sigma _i(r_1,\ldots ,r_d)\left( x+\frac{1}{x}\right) ^{d-i}\right) =g_b(x), \end{aligned}$$

where $\sigma _0$ is taken as 1.

Example 10

The set $V_1$ is the closed interval $[-2,2]$. For $d=2$, we have that $(b_1,b_2)=\varPhi (r_1,r_2)=(-r_1-r_2,r_1r_2+2)$ and, as $\varPhi$ transforms the boundary of $I_2$ in the boundary of $V_2$, we have that

$$\begin{aligned} V_2=\{(b_1,b_2): 2|b_1|-2 \le b_2\le \frac{b_1^2}{4}+2\}. \end{aligned}$$

(5)

More generally, a description of the space of complex polynomials such that all of their roots lie in $S^1$ can be given.

Definition 11

If the complex polynomial $L(z)=z^k+c_1z^{k-1}+\cdots +c_{k-1}z+c_k$ satisfies $c_k=1$ and $\overline{c_i}=c_{k-i}$ for $i=1\ldots ,k$, then it is called conjugate reciprocal.

In [11], the geometry, topology and Lebesgue measure of the space of conjugate reciprocal polynomials of fixed degree with all roots in $S^1$ is studied. To state some of their results, we introduce the matrix $X_k\in \mathbb {C}^{k-1\times k-1}$ by giving its l, j entry:

$$\begin{aligned} {[}X_k]_{l,j}=\left\{ \begin{array}{lll} \delta _{l,j}+\delta _{k-l,j} &{} \text {if} &{} 1\le l<k/2 \\ \delta _{l,j} &{} \text {if} &{} l=k/2 \\ i\delta _{k-l,j}-i\delta _{l,j} &{} \text {if} &{} k/2< l<k, \\ \end{array}\right. \end{aligned}$$

(6)

where $\delta _{l,j}$ is the Kronecker delta and i is the imaginary unit (in [11] this matrix is normalized, but this is unimportant for us). The key point here is that the polynomial

$$\begin{aligned} l(x)=(x^k+1)+\sum _{n=1}^{k-1}c_nx^{k-n} \end{aligned}$$

is conjugate reciprocal if and only if there exists $w\in \mathbb {R}^{k-1}$ such that $c=X_kw\in \mathbb {C}^{k-1}$, in which case it is denoted $w(x)=l(x)$. Define $W_k$ as the set

$$\begin{aligned} W_k=\{w\in \mathbb {R}^{k-1}:w(x) \text { has all roots in } S^1\}. \end{aligned}$$

(7)

We have the following theorem which summarizes some of the results in [11]:

Theorem 12

$W_k$ is homeomorphic to the unit ball of $\mathbb {R}^{k-1}$ and its boundary is the set of those $w\in W_k$ such that $\varDelta (w(x))=0$, where $\varDelta$ denotes the discriminant of a given polynomial.

Example 13

To describe the set $W_3$, recall that

$$\begin{aligned} \varDelta (ax^3+bx^2+cx+d)=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd. \end{aligned}$$

Thus, we have

$$\begin{aligned} \varDelta (w_1,w_2)= & {} \varDelta (x^3+(w_1+iw_2)x^2+(w_1-iw_2)x+1)\nonumber \\&=w_1^4+2w_1^2w_2^2+w_2^4-8w_1^3+24w_1w_2^2+18w_1^2+18w_2^2-27 \end{aligned}$$

(8)

and the boundary of $W_3$ is the set $\{(w_1,w_2)\in \mathbb {R}^2:\varDelta (w_1,w_2)=0\}$.

3 2-designs with 7 elements

3.1 Existence

Let $a_1,a_2,a_3\in S^1$ and assume that there is a 2-design X, with $|X|=7$ containing them. This is equivalent to the existence of a polynomial of degree 7

$$\begin{aligned} f(z)=z^7+c_6z^6+\cdots +c_1z+c_0 \end{aligned}$$

(9)

such that

• i) $c_6=c_5=0$ and

• ii) $a_1,a_2,a_3$ are roots of f(z) and the other roots $b_1,b_2,b_3,b_4$ also belong to $S^1$.

As we saw, condition i) can be replaced by $c_2=c_1=0$ and, moreover, these two conditions are equivalent if we assume ii) to be true. Taking this into account, we try to obtain the coefficients of such f(z).

Let $\sigma _1,\sigma _2,\sigma _3$ be the symmetric functions of $a_1,a_2,a_3$ and write

$$\begin{aligned} f(z)=\prod _{i=1}^3(z-a_i)\prod _{i=1}^4(z-b_i)=(z^3-\sigma _1 z^2+\sigma _2 z-\sigma _3)(z^4-\tilde{\sigma _1} z^3+\tilde{\sigma _2} z^2-\tilde{\sigma _3}z+\tilde{\sigma _4}), \end{aligned}$$

(10)

where $\tilde{\sigma _1},\tilde{\sigma _2},\tilde{\sigma _3},\tilde{\sigma _4}$ are the symmetric functions of $b_1,b_2,b_3,b_4$. To obtain the $\tilde{\sigma _i}$’s we impose $c_6=c_5=c_2=c_1=0$. By calculating the product of the two polynomials in the right-hand side of (10) and equating to zero the coefficients of $z^6,z^5,z^2$ and z, we come to

$$\begin{aligned} \tilde{\sigma _1}&=-\sigma _1 \end{aligned}$$

(11)

$$\begin{aligned} \tilde{\sigma _2}&=-\tilde{\sigma _1}\sigma _1-\sigma _2=\sigma _1^2-\sigma _2 \end{aligned}$$

(12)

and to the system

$$\begin{aligned} \left\{ \begin{array}{ccc} \sigma _2\tilde{\sigma _3}+\sigma _1\tilde{\sigma _4} &{} = &{} -\sigma _3(\sigma _1^2-\sigma _2) \\ \sigma _3\tilde{\sigma _3}+\sigma _2\tilde{\sigma _4} &{} = &{} 0 \end{array} \right. \end{aligned}$$

(13)

Assuming that $\sigma _2^2-\sigma _3\sigma _1\ne 0$, we obtain

$$\begin{aligned} \begin{array}{lcr} \tilde{\sigma _3}=-\dfrac{\sigma _3\sigma _2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3},&\,&\tilde{\sigma _4}=\dfrac{\sigma _3^2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3}. \end{array} \end{aligned}$$

(14)

Thus we have obtained the following proposition, to be generalized in Sect. 4.

Proposition 14

Let $a_1,a_2,a_3\in S^1$ such that $\sigma _2^2-\sigma _3\sigma _1\ne 0$.

• i) If the roots $b_1,b_2,b_3,b_4$ of the polynomial

  $$\begin{aligned} G(z):=z^4+\sigma _1z^3+(\sigma _1^2-\sigma _2)z^2+\frac{\sigma _3\sigma _2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3}z+\frac{\sigma _3^2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3} \end{aligned}$$

  (15)

  are unit-norm complex numbers then the set $X=\{a_1,a_2.a_3,b_1,b_2,b_3,b_4\}$ is the unique spherical 2-design with 7 elements that contains $a_1,a_2,a_3$.

• ii) If some root of G(z) does not belong to $S^1$ then there is not a spherical 2-design with 7 elements containing $a_1,a_2,a_3$.

Remark 15

Note that the condition on the polynomial G(z) is actually a condition on $a_1,a_2,a_3$. We specify this in Definition 25 below.

Now, we can apply the results of Sect. 2 to establish when our 2-design actually exists. Surprisingly, the polynomial G(z) turns out to be self-inversive with no further conditions on the numbers $a_1,a_2, a_3\in S^1$ (this is easy to check using $\bar{\sigma _2}\sigma _3=\sigma _1$ and $|\sigma _3|=1$, see Proposition 31 below for the general case), so we apply Proposition 6 to $G'(z)$ in order to satisfy the second condition of Theorem 5.

Corollary 16

Let $a_1,a_2,a_3\in S^1$ such that $\sigma _2^2-\sigma _3\sigma _1\ne 0$ and G(z) the polynomial defined in Eq.(15). In the following cases there exists a 2-design containing them.

• 1. If $2\sigma _3\sigma _2\ne 3\sigma _1(\sigma _2^2-\sigma _3\sigma _1)$, $0<|\sigma _1|<\frac{4}{3}$ and $\left| 3\sigma _1\sigma _2^2-3\sigma _1^2\sigma _3-2\sigma _2\sigma _3\right| <8-\frac{9}{2}|\sigma _1|^2-\left| 4\sigma _1\sigma _3-\frac{5}{2}\sigma _2^2\right|$.

• 2. If $2\sigma _3\sigma _2= 3\sigma _1(\sigma _2^2-\sigma _3\sigma _1)$ and $\frac{3}{4}\sigma _1,\dfrac{\sigma _1^2-\sigma _2}{2}\in D$.

• 3. If $|\sigma _1|<\alpha$, where $\alpha$ is the real root of the polynomial $3x^3+10x^2+6x-8$ ($\alpha \simeq 0.60712$).

Proof

The first two cases follow from applying Proposition 6 to

$$\begin{aligned} \frac{1}{4}G'(z)=z^3+\frac{3}{4}\sigma _1z^2+\frac{\sigma _1^2-\sigma _2}{2}z+\frac{\sigma _3\sigma _2(\sigma _1^2-\sigma _2)}{4(\sigma _2^2-\sigma _1\sigma _3)}, \end{aligned}$$

using the fact that $|\sigma _1^2-\sigma _2|=|\sigma _2^2-\sigma _1\sigma _3|$, which follows from $\sigma _3^2\overline{(\sigma _1^2-\sigma _2)}=\sigma _2^2-\sigma _1\sigma _3$ (note also that the case $\sigma _1=0$ is excluded because, due to $\bar{\sigma _2}\sigma _3=\sigma _1$, it would implies $\sigma _2=0$ and then $\sigma _2^2-\sigma _3\sigma _1=0$). The condition of the third case is stronger than the two previous ones. If $2\sigma _3\sigma _2\ne 3\sigma _1(\sigma _2^2-\sigma _3\sigma _1)$, apply triangle inequality and, if $2\sigma _3\sigma _2= 3\sigma _1(\sigma _2^2-\sigma _3\sigma _1)$, the two inequalities follows. $\square$

Remark 17

For arbitrary t, we will define a general polynomial $G_{k,t}$ (see Definition 25 below) so that $G(z)=G_{3,2}(z)$ and obtain a general condition implying that all of its roots are in the unit circle (see Corollary 32). Nevertheless, in the case $t=2$, the conditions given by Corollary 16 are more accurate.

We can use Example 10 to give a family of examples of 2-designs.

Example 18

Let $z_0\in S^1$ and set

$$\begin{aligned} \begin{array}{ccc} a_1=1, &{} a_2=z_0, &{} a_3=\overline{z_0}. \\ \end{array} \end{aligned}$$

We want to find out when there is a 2-design containing $1,z_0,\overline{z_0}$. We have

$$\begin{aligned} \begin{array}{cc} \sigma _1=\sigma _2=1+2\mathfrak {R}(z_0), &{} \sigma _3=1, \\ \end{array} \end{aligned}$$

then $\sigma _2^2-\sigma _1\sigma _3\ne 0$ if and only if $z_0\ne \pm i$ and $z_0\ne -\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$, in order to be able to define the polynomial in (15), we discard these cases.

Thus, we need to check when the polynomial

$$\begin{aligned} G_0(z)= z^4+(1+2\mathfrak {R}(z_0))z^3+2\mathfrak {R}(z_0)(1+2\mathfrak {R}(z_0))z^2+(1+2\mathfrak {R}(z_0))z+1 \end{aligned}$$

(16)

has all of its roots in $S^1$. Note first that 1 is not a root of $G_0$, because in that case $\mathfrak {R}(z_0)\notin \mathbb {R}$, and if $-1$ is a root, has multiplicity 2. By (5), taking $b_1=1+2\mathfrak {R}(z_0)$ and $b_2=2\mathfrak {R}(z_0)(1+2\mathfrak {R}(z_0))$, we have that $G_0(z)$ has all of its roots in $S^1$ if and only if $\mathfrak {R}(z_0)$ belongs to the set

$$\begin{aligned} \left\{ x\in \mathbb {R}:2|2x+1|-2\le 2x(2x+1)\le \tfrac{(2x+1)^2}{4}+2\right\} =\left[ \tfrac{-1-\sqrt{28}}{6},0\right] \cup \left[ \tfrac{1}{2},\tfrac{-1+\sqrt{28}}{6}\right] . \end{aligned}$$

As $\tfrac{-1-\sqrt{28}}{6}\simeq -1,0485$ and $z_0\in S^1$, we conclude that $G_0(z)$ has all of its roots in $S^1$ if and only if $\mathfrak {R}(z_0)\in \left[ -1,-\frac{1}{2}\right) \cup \left( -\frac{1}{2},0\right) \cup \left[ \tfrac{1}{2},\tfrac{-1+\sqrt{28}}{6}\right]$. We depict, for some particular values of $z_0$, the corresponding 2-designs in Fig. 1. The grey points are the roots of the corresponding $G_0(z)$.

Remark 19

Notice that, in the previous example, the condition $\sigma _2^2-\sigma _1\sigma _3=0$ implies that our three input points are either the cubic roots of the unity or three of the four quartic roots of the unity. Therefore, in both cases there exists an infinite number of group-type 2-designs containing them. It is also worth noticing that the cases $z_0=-1$ and $z_0=\tfrac{1}{2}$ both correspond to the group-type 2-design given by the union of the quartic roots of 1 and the cubic roots of $-1$. Here $-1$ belongs to both sets of roots.

Remark 20

If we would have applied Corollary 16 to work this example, we would have obtained that for $\mathfrak {R}(z_0)\in \left( -1,-\frac{1}{2}\right) \cup \left( -\frac{1}{2},0\right)$ there exists a 2-design containing $1,z_0,\overline{z_0}$, that is, a less precise condition.

Fig. 1

Examples of 2-designs

3.2 Characterization of 2-designs

We would like to give a reciprocal statement to Proposition 14: Does any 2-design with 7 elements can be obtained in this way? To answer to this question we need the following lemmas.

Lemma 1

Let $a_1,a_2,a_3\in S^1$ such that $\sigma _2^2-\sigma _3\sigma _1=0$. Then one of the following is true:

• i) $a_1,a_2,a_3$ are the cubic roots of a unit-norm complex number, or

• ii) some pair of the $a_i$’s are opposite.

Proof

As the polynomial

$$\begin{aligned} \prod _{i=1}^3(z-a_i)=z^3-\sigma _1 z^2+\sigma _2 z-\sigma _3 \end{aligned}$$

is self inversive, we have $\overline{\sigma _1}\sigma _3=\sigma _2$ and $|\sigma _3|=1$. Then $|\sigma _1|^2=|\sigma _2|^2=|\sigma _1|$, that is $|\sigma _1|=0$ or $|\sigma _1|=1$. If $|\sigma _1|=0$ we have $\sigma _1=\sigma _2=0$ and then $a_1,a_2,a_3$ are the cubic roots of $\sigma _3$. On the other hand, if $|\sigma _1|=1$, we have $\sigma _3=\sigma _1\overline{\sigma _1}\sigma _3=\sigma _1\sigma _2$ and then $a_1,a_2,a_3$ are the roots of

$$\begin{aligned} z^3-\sigma _1 z^2+\sigma _2 z-\sigma _1\sigma _2=(z-\sigma _1)(z^2+\sigma _2) \end{aligned}$$

(17)

which implies ii). $\square$

Lemma 2

Let X be a finite subset of $S^1$ such that $|X|\ge 3$ and for any three distinct points in X, $\sigma _2^2-\sigma _3\sigma _1=0$, where $\sigma _1,\sigma _2,\sigma _3$ are the corresponding symmetric functions. Then one the following options is true:

• (i) $|X|=3$ and the elements of X are the cubic roots of a unit-norm complex number, or

• (ii) $|X|=4$ and the elements of X are the quartic roots of a unit-norm complex number.

Proof

If there exists $a_1,a_2,a_3$ distinct elements of X such that $\sigma _1=0$, then, by Lemma 1, they are the cubic roots of $\sigma _3$ and, moreover, there are not other (distinct) elements in X. In fact, let $a_4\in X$ such that $a_4\ne a_i$, $i=1,2,3$. Applying Lemma 1 to $a_1,a_2,a_4$ we have that $a_4=-a_1$ or $a_4=-a_2$. If $a_4=-a_1$ apply Lemma 1 to $a_3,a_2,a_4$ and get $a_1=a_2$ or $a_1=a_3$, a contradiction. Similarly if $a_4=-a_2$. We conclude that $a_1,a_2,a_3$ are the only distinct elements of X.

So, we are left with the case of $|\sigma _1|=1$: take three distinct points, they are the roots of a polynomial as in (17). As we have $\sigma _2=\sigma _1^2$, any other distinct point in X must be equal to $-\sigma _1$ so the elements of X are the roots of $z^4-\sigma _1^4$. $\square$

Corollary 21

Let X be a spherical 2-design with 7 elements, and at least five of them are distinct. Then there exists three elements of X such that $\sigma _2^2-\sigma _3\sigma _1\ne 0$, where $\sigma _1,\sigma _2,\sigma _3$ are the corresponding symmetric functions.

Now we can apply Proposition 14 to obtain our characterization result.

Proposition 22

Let X a 2-design with 7 elements, and at least five of them distinct. Then there exist $a_1,a_2,a_3\in X$ such that $X=\{a_1,a_2.a_3,b_1,b_2,b_3,b_4\}$ where $b_1,b_2,b_3,b_4$ are the roots of the polynomial G(z) defined in (15) and $\sigma _1,\sigma _2,\sigma _3$ are the symmetric functions of $a_1,a_2,a_3$.

3.3 Group-type 2-designs

What can be said about group-type 2-designs under this point of view? To answer that question, first note the following fact, whose proof is evident.

Proposition 23

A spherical 2-design of 7 elements is of group-type if and only if it is given by the roots of a polynomial of the shape

$$\begin{aligned} f(z)=z^7+\alpha z^4+\beta z^3+\gamma \end{aligned}$$

(18)

where one of the following options holds:

• i) $\alpha ,\beta \in S^1$ and $\gamma =\alpha \beta$.

• ii) $\alpha =\beta =0$ and $\gamma \in S^1$.

We would like to describe group-type t-designs in terms of symmetric functions. The following proposition shows that this is possible.

Proposition 24

Let $a_1,a_2,a_3\in S^1$ such that $\sigma _2^2-\sigma _3\sigma _1\ne 0$. If $\sigma _3=\sigma _2\sigma _1$ then there is a spherical 2-design X containing $a_1,a_2,a_3$. Moreover, X is a group-type 2-design.

Proof

We know by Proposition 14 that if there is a 2-design with 7 elements containing $a_1,a_2,a_3$, their other four elements must be the roots of the polynomial G(z) in equation (15), provided that all of them are in $S^1$. Instead of finding this out, we make the product

$$\begin{aligned} f(z)=(z^3-\sigma _1 z^2+\sigma _2 z-\sigma _3)\left( z^4+\sigma _1z^3+(\sigma _1^2-\sigma _2)z^2+\frac{\sigma _3\sigma _2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3}z+\frac{\sigma _3^2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3}\right) \end{aligned}$$

(19)

and show that f(z) satisfies the first condition on Proposition 23.

First, observe that, due to $a_1,a_2,a_3\in S^1$, we have $\overline{\sigma _1}\sigma _3=\sigma _2$, which, together with our assumption, gives

$$\begin{aligned} |\sigma _1|^2\sigma _2=\sigma _2. \end{aligned}$$

If $\sigma _2=0$ then $\sigma _1=0$ (because $|\sigma _3|=1$), which would imply $\sigma _2^2-\sigma _3\sigma _1=0$, a contradiction. So $|\sigma _1|=|\sigma _2|=1$. We also have,

$$\begin{aligned} \sigma _2^2-\sigma _1\sigma _3=\sigma _2(\sigma _2-\sigma _1^2) \end{aligned}$$

and then

$\begin{array}{cclc} f(z) &{} = &{} (z^3-\sigma _1 z^2+\sigma _2 z-\sigma _1\sigma _2)(z^4+\sigma _1z^3+(\sigma _1^2-\sigma _2)z^2-\sigma _1\sigma _2z-\sigma _1^2\sigma _2) &{} = \\ &{} = &{} z^7-\sigma _1^3z^4-\sigma _2^2 z^3+\sigma _1^3\sigma _2^2. &{} \\ \end{array}$ $\square$

Alternatively, we can combine Propositions 23, 14 and 22 to obtain necessary and sufficient conditions for 2-design with 7 elements to be of group-type . In fact, both two cases in Proposition 23 applied to the polynomial

$$\begin{aligned} \left( z^3-\sigma _1 z^2+\sigma _2 z-\sigma _3\right) \left( z^4+\sigma _1z^3+(\sigma _1^2-\sigma _2)z^2+\frac{\sigma _3\sigma _2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3}z+\frac{\sigma _3^2(\sigma _1^2-\sigma _2)}{\sigma _2^2-\sigma _1\sigma _3}\right) \end{aligned}$$

(20)

turn out in algebraic and norm conditions on the $\sigma _i$’s, though these conditions will be rather cumbersome. In Sect. 5 further tools to understand the space of t-designs in the $\sigma$-space will be given.

4 Spherical t-designs in ${\mathbb {R}}^2$

4.1 Existence

In this section we generalize the principal results obtained for the case $t=2$. To state our main theorem we need the following definition.

Definition 25

Let t, k be positive integer numbers. Let $a_1,\ldots ,a_k\in S^1$ and $\sigma _1,\ldots ,\sigma _k$ their corresponding symmetric functions. We say that $a_1,\ldots ,a_k$ are in t-good position if the following two conditions hold:

• i) The determinant $\varDelta _{k,t}=\varDelta _{k,t}(\sigma _1,\ldots ,\sigma _k)$ of the matrix

  $$\begin{aligned} T_{k,t}=\left[ \begin{array}{ccccll} \sigma _{k-1} &{} \sigma _{k-2} &{} \sigma _{k-3} &{} \dots &{} \sigma _{k-t+1} &{} \sigma _{k-t} \\ \sigma _{k} &{} \sigma _{k-1} &{} \sigma _{k-2} &{} \dots &{} \sigma _{k-t+2} &{} \sigma _{k-t+1} \\ 0 &{} \sigma _{k} &{} \sigma _{k-1} &{} \dots &{} \sigma _{k-t+3} &{} \sigma _{k-t+2} \\ 0 &{} 0 &{} \sigma _k &{} \dots &{} \sigma _{k-t+4} &{} \sigma _{k-t+3} \\ \vdots &{} \vdots &{} \dots &{} \dots &{} \dots &{} \vdots \\ 0 &{} 0 &{} \dots &{} 0 &{}\sigma _k &{} \sigma _{k-1} \\ \end{array} \right] \in \mathbb {C}^{t\times t} \end{aligned}$$

  (21)

  is non-zero. Here $\sigma _i$ is taken as zero if $i\le 0$.

• ii) All of the roots of the polynomial

  $$\begin{aligned} G_{k,t}(z)= z^{2t}-\tilde{\sigma _1}z^{2t-1}+\tilde{\sigma _2}z^{2t-2}-\dots -\tilde{\sigma }_{2t-1}z+\tilde{\sigma }_{2t} \end{aligned}$$

  (22)

  lie in $S^1$, where the coefficients $\tilde{\sigma }_i$ are defined as follows: for $i=1,\ldots ,t$ they satisfy the recursion formula

  $$\begin{aligned} \begin{array}{cccc} \tilde{\sigma }_1=-\sigma _1&\text {and}&\tilde{\sigma }_i=-\sigma _i-\sum _{j=1}^{i-1}\tilde{\sigma }_j\sigma _{i-j}&\text {for } i=2,\ldots ,t \end{array} \end{aligned}$$

  (23)

  while, for $i=t+1,\ldots ,2t$, they are obtained as the unique solutions of the system

  $$\begin{aligned} T_{k,t}\cdot \left[ \begin{array}{c} \tilde{\sigma }_{t+1} \\ \tilde{\sigma }_{t+2} \\ \vdots \\ \tilde{\sigma }_{2t} \\ \end{array}\right] =\left[ \begin{array}{c} -\sigma _k\tilde{\sigma }_{t} \\ 0 \\ \vdots \\ 0 \end{array} \right] \end{aligned}$$

  (24)

Example 26

For any t the matrix $T_{1,t}$ is singular, so one single point is never in t-good position.

Example 27

Equations (11),(12) and (14) give the expressions for the $\tilde{\sigma _i}'$s if $t=2$, $k=3$ and $a_1,a_2,a_3\in S^1$ are such that $\varDelta _{3,2}=\sigma _2^2-\sigma _1\sigma _3\ne 0$. The polynomial $G_{3,2}(z)$ is equal to the polynomial G(z) in (15).

Example 28

Set $t=3$ and $k=4$. The expressions for $\tilde{\sigma }_i$ are, for $i=1,2,3:$

$$\begin{aligned} \begin{array}{ccc} \tilde{\sigma }_1=-\sigma _1,&\tilde{\sigma }_2=-\sigma _2+\sigma _1^2,&\tilde{\sigma }_3=-\sigma _3+2\sigma _1\sigma _2-\sigma _1^3. \end{array} \end{aligned}$$

Assuming $\varDelta _{4,3}=\sigma _3^3-2\sigma _2\sigma _3\sigma _4+\sigma _1\sigma _4^2\ne 0$, we can solve the system

$$\begin{aligned} \left[ \begin{array}{ccc} \sigma _{3} &{} \sigma _{2} &{} \sigma _1 \\ \sigma _{4} &{} \sigma _{3} &{} \sigma _2 \\ 0 &{} \sigma _{4} &{} \sigma _3 \\ \end{array} \right] \cdot \left[ \begin{array}{c} \tilde{\sigma }_{4} \\ \tilde{\sigma }_{5} \\ \tilde{\sigma }_{6} \\ \end{array}\right] =\left[ \begin{array}{c} -\sigma _4\tilde{\sigma }_{3} \\ 0 \\ 0 \\ \end{array} \right] \end{aligned}$$

to obtain

$$\begin{aligned} \tilde{\sigma }_4= & {} \dfrac{\sigma _4(\sigma _3-2\sigma _1\sigma _2+\sigma _1^3)(\sigma _3^2-\sigma _4\sigma _2)}{\varDelta _{4,3}},\\ \tilde{\sigma }_5= & {} \dfrac{-\sigma _4^2\sigma _3(\sigma _3-2\sigma _1\sigma _2+\sigma _1^3)}{\varDelta _{4,3}},\\ \tilde{\sigma }_6= & {} \dfrac{\sigma _4^3(\sigma _3-2\sigma _1\sigma _2+\sigma _1^3)}{\varDelta _{4,3}}. \end{aligned}$$

Our main theorem below, allows to find a t-design with $2t+k$ elements, given k points in t-good position.

Theorem 29

For k complex numbers in $S^1$ in t-good position, there is an unique t-design X containing them with $|X|=k+2t$.

Proof

Assume that $a_1,\ldots ,a_k\in S^1$ are in t-good position and let

$$\begin{aligned} L(z)=\prod _{i=1}^k(z-a_i)=z^k+\sum _{i=1}^{k}(-1)^i\sigma _iz^{k-i}. \end{aligned}$$

(25)

By hypothesis, the set X of the roots of the polynomial

$$\begin{aligned} L(z)G_{k,t}(z)=z^n+\sum _{i=1}^{n}c_{n-i} z^{n-i}, \end{aligned}$$

where $n=k+2t$, is a subset of $S^1$. Thus, to show that X is a spherical t-design it only suffices to check that $c_{n-1}=c_{n-2}=\cdots =c_{n-t}=0$ and this is clear for the construction of the coefficients of $G_{k,t}(z)$.

The unicity also follows from the construction of $G_{k,t}(z)$ because, if $a_1,\ldots ,a_k$ belongs to a spherical t-design X with $n=k+2t$ elements, and $\{b_1,\ldots ,b_{n-k}\}=X\setminus \{a_1,\ldots ,a_k\}$ then the $b_i$’s are the roots of a polynomial

$$\begin{aligned} H(z)=z^{2t}+\sum _{i=1}^{2t}(-1)^i\phi _iz^{2t-i}, \end{aligned}$$

where the $\phi _i$’s are the symmetric functions of $b_1,\ldots ,b_{n-k}$. Being X a spherical t-design, the polynomial

$$\begin{aligned} L(z)H(z)=z^n+\sum _{i=1}^{n}d_{n-i} z^{n-i}, \end{aligned}$$

must satisfy $d_{n-i}=d_i=0$, $i=1,\ldots ,t$ and this implies that $\phi _i=\tilde{\sigma _i}$ for $i=1,\ldots ,2t$, where the $\tilde{\sigma _i}$’s are as in Definition 25. Thus, $b_1,\ldots ,b_{n-k}$ are the roots of $G_{k,t}(z)=H(z)$. $\square$

Remark 30

Assume $t\ge 2$. From Theorems 29 and 3 we conclude that if 2 points in $S^1$ are in t-good position then there exists a unique group-type t-design (the union of two $t+1$-gons) containing them.

In order to construct t-designs, we need to decide, for k complex numbers in $S^1$ such that $\varDelta _{k,t}\ne 0$, whether the polynomial $G_{k,t}(z)$ have all of its roots in $S^1$. This can be done using Theorem 5. Unexpectedly, $G_{k,t}(z)$ satisfies the first one of the conditions of the theorem if $k\ge t+1$.

Proposition 31

Let t, k be positive integers such that $k\ge t+1$. If $a_1,\ldots ,a_k\in S^1$ satisfy that $\varDelta _{k,t}\ne 0$, then the polynomial $G_{k,t}(z)$ is self-inversive.

Proof

As $G_{k,t}(z)$ is monic, we need to show that the constant term $\tilde{\sigma }_{2t}$ is a unit-norm complex number and that $\tilde{\sigma }_{2t}\overline{\tilde{\sigma }_i}=\tilde{\sigma }_{2t-i}$, for $i=1,\ldots ,t$. Note that, by Cramer’s rule,

$$\begin{aligned} \tilde{\sigma }_{2t}=\dfrac{(-1)^{t}\sigma ^t_k\tilde{\sigma _t}}{\varDelta _{k,t}}. \end{aligned}$$

(26)

As $|\sigma _k|=1$, it is enough to prove that $|\tilde{\sigma _t}|=|\varDelta _{k,t}|$. We will show that, for any t,

$$\begin{aligned} \sigma _k^t\overline{\varDelta _{k,t}}=(-1)^t\tilde{\sigma _t}, \end{aligned}$$

(27)

which implies the previous assertion. Observe that, if $t=1$, $\sigma _k\overline{\varDelta _{k,1}}=\sigma _k\overline{\sigma _{k-1}}=\sigma _1=-\tilde{\sigma _1}$ and that if $t=2$, $\sigma _k^2\overline{\varDelta _{k,2}}=\sigma _k^2\overline{(\sigma _{k-1}^2-\sigma _k\sigma _{k-2})}=\sigma _1^2-\sigma _2=\tilde{\sigma _2}$. So, suppose that $t\ge 3$ and that (27) is true for lower values of t. We can express $\varDelta _{k,t}$ as

$$\begin{aligned} \varDelta _{k,t}=\sum _{i=1}^{t-2}(-1)^{i-1}\sigma _k^{i-1}\sigma _{k-i}\varDelta _{k,t-i}+(-1)^{t-2}\sigma _k^{t-2}(\sigma _{k-t+1}\sigma _{k-1}-\sigma _{k-t}\sigma _k). \end{aligned}$$

(28)

Then,

$$\begin{aligned} \sigma _k^t\overline{\varDelta _{k,t}} =&\sum _{i=1}^{t-2}(-1)^{i-1}\sigma _i\sigma _k^{t-i}\overline{\varDelta _{k,t-i}}+(-1)^{t-2}\sigma _k^2\overline{(\sigma _{k-t+1}\sigma _{k-1}-\sigma _{k-t}\sigma _k)}=\\ =&\sum _{i=1}^{t-2}(-1)^{i-1}\sigma _i\tilde{\sigma }_{t-i}(-1)^{t-i}+(-1)^{t-2}(\sigma _{t-1}\sigma _1-\sigma _t)= \\ =&(-1)^t\left( -\sum _{i=1}^{t-2}\sigma _i\tilde{\sigma }_{t-i}-\sigma _{t-1}\tilde{\sigma _1}-\sigma _t\right) =(-1)^t\tilde{\sigma _t}. \end{aligned}$$

We now prove that

$$\begin{aligned} \tilde{\sigma }_{2t}\overline{\tilde{\sigma }_i}=\tilde{\sigma }_{2t-i}, \end{aligned}$$

(29)

for $i=1,\ldots ,t-1$. If $i=1$,

$$\begin{aligned} \tilde{\sigma }_{2t}\overline{\tilde{\sigma }_1}=-\tilde{\sigma }_{2t}\overline{\sigma _1}=\dfrac{(-1)^{t-1}\sigma _{k-1}\sigma _k^{t-1}\tilde{\sigma }_t}{\varDelta _{k,t}} \end{aligned}$$

and this is easily seen to be equal to $\tilde{\sigma }_{2t-1}$ using, again, Cramer’s rule. Now, take $1<i\le t-1$ and assume that $\tilde{\sigma }_{2t}\overline{\tilde{\sigma }_j}=\tilde{\sigma }_{2t-j}$ holds for $j<i$. By (24) we have that

$$\begin{aligned} \sum _{j=0}^{i}\sigma _{k-j}\tilde{\sigma }_{2t-(i-j)}=0. \end{aligned}$$

Thus,

$$\begin{aligned} \sigma _k\tilde{\sigma }_{2t-i}=-\sum _{j=1}^{i}\sigma _{k-j}\tilde{\sigma }_{2t-(i-j)}=-\sum _{j=1}^{i}\sigma _{k-j}\tilde{\sigma }_{2t}\overline{\tilde{\sigma }_{i-j}}, \end{aligned}$$

and we finally obtain

$$\begin{aligned} \tilde{\sigma }_{2t-i}=-\tilde{\sigma }_{2t}\sum _{j=1}^{i}\overline{\sigma _k}\sigma _{k-j}\overline{\tilde{\sigma }_{i-j}}=-\tilde{\sigma }_{2t}\overline{\sum _{j=1}^{i}\sigma _{j}\tilde{\sigma }_{i-j}}=\tilde{\sigma }_{2t}\overline{\tilde{\sigma }_i}. \end{aligned}$$

The only thing left to prove to conclude is $\tilde{\sigma }_{2t}\overline{\tilde{\sigma }_t}=\tilde{\sigma }_t$, but this is straightforward, because we have $\tilde{\sigma }_{2t}\varDelta _{k,t}=(-1)^t\sigma _k^t\tilde{\sigma _t}$ and from (27) it follows easily that

$$\begin{aligned} \varDelta _{k,t}=(-1)^t\sigma _k^t\overline{\tilde{\sigma _t}} \end{aligned}$$

(30)

which finishes the proof. $\square$

Finding under what conditions on $a_1,\ldots ,a_k$ the polynomial $G_{k,t}(z)$ satisfies the second condition of Cohn’ Theorem could be a difficult task and we do not have at our disposal a result as Corollary 16 for $t>2$. Instead of that, we can apply Theorem 7 to arrive to sufficient conditions for the existence of t-designs.

Corollary 32

Let t, k be positive integers such that $k\ge t+1$. If $a_1,\ldots ,a_k\in S^1$ satisfy that $\varDelta _{k,t}\ne 0$, and that

$$\begin{aligned} \sum _{i=1}^{t-1}2|\tilde{\sigma }_i|+|\tilde{\sigma }_t|\le 2 \end{aligned}$$

(31)

then $a_1,\ldots ,a_k$ are in t-good position. Consequently, Theorem 29 guarantees the existence and unicity of a t-design X, with $|X|=k+2t$, containing them.

Proof

As we saw, $|\tilde{\sigma }_{2t}|=1$ and $\tilde{\sigma }_{2t}\overline{\tilde{\sigma }_i}=\tilde{\sigma }_{2t-i}$ for $i=1,\ldots ,t-1$, which implies that $|\tilde{\sigma }_i|=|\tilde{\sigma }_{2t-i}|$ for $i=1,\ldots ,t-1$. Then, Theorem 7 applied to the monic polynomial $G_{k,t}$ gives the result. $\square$

Example 33

Applying Corollary 32 to the case $t=2,k=3$, we obtain that the condition

$$\begin{aligned} 2|\sigma _1|+|\sigma _1^2-\sigma _2|\le 2 \end{aligned}$$

(32)

assures the existence and unicity of a 2-design containing $a_1,a_2,a_3\in S^1$ such that $\sigma _2^2-\sigma _1\sigma _3\ne 0$. If we set $a_1=z_0$, $a_2=\overline{z_0}$ and $a_3=1$ with $z_0\in S^1$ we obtain the existence of 2-designs for $\tfrac{1-\sqrt{17}}{4}\simeq -0,78077\le \mathfrak {R}(z_0)<-\tfrac{1}{2}$. This comes from solving the equation

$$\begin{aligned} 2|1+2\mathfrak {R}{z_0}|+|(1+2\mathfrak {R}{z_0})^2-(1+2\mathfrak {R}{z_0})|\le 2. \end{aligned}$$

Compare this with Example 18 and with Remark 20.

Example 34

In this example, we construct a family of 3-designs in a similar fashion than we did in Example 18. In order to apply Corollary 32, we must take 4 points in $S^1$ such that $\varDelta _{4,3}\ne 0$.

Let $z_0,z_1\in S^1$ and set

$$\begin{aligned} \begin{array}{cccc} a_1=z_0, &{} a_2=\overline{z_0}, &{} a_3=z_1, &{} a_4=\overline{z_1}. \\ \end{array} \end{aligned}$$

We have

$$\begin{aligned} \begin{array}{ccc} \sigma _1=\sigma _3=2(\mathfrak {R}(z_0)+\mathfrak {R}(z_1)), &{} \sigma _2=2+2(\mathfrak {R}(z_0z_1)+\mathfrak {R}(z_0\overline{z_1})), &{} \sigma _4=1. \\ \end{array} \end{aligned}$$

Based on Example 28, we obtain the expressions for the $\tilde{\sigma _i}'s$. We put $z_0=e^{i\theta _0}$ and $z_1=e^{i\theta _1}$ so that $\mathfrak {R}(z_0)=\cos \theta _0$ and $\mathfrak {R}(z_1)=\cos \theta _1$ and we obtain:

$$\begin{aligned} \begin{array}{ccc} \tilde{\sigma _1}=\tilde{\sigma _5}=-2(\cos \theta _0+\cos \theta _1),&\tilde{\sigma _2}=\tilde{\sigma _4}=4\cos ^2\theta _0+4\cos ^2\theta _1+4\cos \theta _0\cos \theta _1-2,&\tilde{\sigma _6}=1. \end{array} \end{aligned}$$

whereas

$$\begin{aligned} \varDelta _{4,3}=-\tilde{\sigma _3}=2(\cos \theta _0+\cos \theta _1)(4\cos ^2\theta _0+4\cos ^2\theta _1-3) \end{aligned}$$

Notice that we could obtain, through Proposition 8 conditions for $G_{4,3}(z)=\sum _{i=0}^6(-1)^i\tilde{\sigma _i}z^{6-i}\in \mathbb {R}[z]$ to have all of its roots lying in $S^1$, however this will lead us to too many calculations. Moreover, even an application of Corollary 32 in the general case would carry too much work to obtain a condition on $\theta _0$ and $\theta _1$. So, instead of doing that, we put $\theta _0=\frac{\pi }{3}$ and look for those $\theta _1$ such that condition (31) holds. This will give us a family of 3-designs. Writing $\theta$ instead of $\theta _1$, condition (31) is equivalent to

$$\begin{aligned} |1+2\cos \theta |+|4\cos ^2\theta +2\cos \theta -1|+|1+2\cos \theta ||2\cos ^2\theta -1|\le 1. \end{aligned}$$

(33)

which holds for $-0.82948...\le \cos \theta \le -0.61803...$. Then, for the corresponding values of $\theta$, we can assure the existence of 3-designs. The corresponding design for $\cos \theta _1=-\frac{3}{4}$ is depicted in Fig. 2 on the left. As in Example 33, conditions for existence are not sharp: on the right of Fig. 2 there is a 3-design whose existence is not provided by the calculations in this example, but obtained in the same fashion.

Fig. 2

Examples of 3-designs

It is clear that a rotation applied to all elements of a t-design gives another t-design. So, we expect our method for constructing t-designs from a subset of $S^1$ will produce a consistent result if we rotate that subset.

Proposition 35

Let $a_1,\ldots ,a_k$ be in t-good position and X the t-design containing them. If $\theta \in \mathbb {R}$ then $e^{i\theta }a_1,\ldots ,e^{i\theta }a_k$ are in t-good position and the corresponding t-design $X_{\theta }$ is obtained from X by a rotation in the angle $\theta$.

Proof

We claim that the coefficient of $z^j$ of the polynomial $G_{k,t}(z)$ is a homogeneous function of degree $2t-j$ when considered as a function on $a_1,\ldots ,a_k$. Thus, denoting $G_{\theta }(z)$ the corresponding polynomial for $e^{i\theta }a_1,\ldots ,e^{i\theta }a_k$, we have that $G_{\theta }(z)=e^{i2t\theta }G_{k,t}(e^{-i\theta }z)$. Then, the roots of $G_{\theta }$ are of the form $e^{i\theta }b$ where b is a root of $G_{k,t}$.

We now prove the claim. For $j=1,\ldots ,t$, it is clear by definition (cf.(23)) that the symmetric functions $\tilde{\sigma _i}$ are homogeneous of degree j. For $j=t+1,\ldots ,2t$, note first that $\varDelta _{k,t}$ is homogeneous of degree $tk-t$ (considered as a function on $a_1,\ldots ,a_k$). In fact, from (27), we easily obtain $\varDelta _{k,t}=(-1)^t\sigma _k^t\overline{\tilde{\sigma _t}}$ (note that this also implies that $\varDelta _{k,t}\ne 0$ when applied to $e^{i\theta }a_1,\ldots ,e^{i\theta }a_k$).

Now, from (26), $\tilde{\sigma }_{2t}$ is homogeneous of degree $tk+t-(tk-t)=2t$ and finally, (29) concludes the proof. $\square$

It would be nice to have a result like Proposition 22 in the general case. Even in the case $t=3$, $k=4$ it is not clear what implies for 4 points in $S^1$ to satisfy $\varDelta _{4,3}=\sigma _3^3-2\sigma _2\sigma _3\sigma _4+\sigma _1\sigma _4^2=0$. In particular, any 4 of the quintic roots of a unit-norm complex number fulfill this condition and it is clear that they belong to an infinite number of 3-designs with 10 elements. Nevertheless, we believe that the following statement is true:

Conjecture 36

Every t-design of $2t+k$ elements can be obtained from k points in t-good position using the methods developed here.

4.2 Group-type t-designs

In the general case, it is also possible to describe group-type spherical t-designs in the $\sigma$-space. We focus on the case $k=t+1$. The following proposition generalizes Proposition 24.

Proposition 37

Let $k=t+1$ and $a_1,\ldots ,a_k\in S^1$ such that $\varDelta _{k,t}\ne 0$. If $\sigma _k=\sigma _1\sigma _{k-1}$ and $\sigma _i=0$ for $i\ne 1,k-1,k$ then $a_1,\ldots ,a_k$ are in t-good position and the t-design containing them is of group-type.

Proof

As before, $\sigma _k=\sigma _1\sigma _{k-1}$ is equivalent to $|\sigma _1|=|\sigma _{k-1}|=1$. Then, it is enough to show that

$$\begin{aligned} L(z)G_{k,t}(z)=(z^{2t}-\sigma _{k-1}^2)(z^{k}-\sigma _1^k), \end{aligned}$$

where L(z) is as in (25). In this case

$$\begin{aligned} L(z)=z^k-\sigma _1z^{k-1}+(-1)^{k-1}\sigma _{k-1}z+(-1)^k\sigma _1\sigma _{k-1}. \end{aligned}$$

In order to calculate $G_{k,t}$, we use (23) to obtain

$$\begin{aligned} \tilde{\sigma _i}=(-1)^i\sigma _1^i, \end{aligned}$$

(34)

for $i=1,\ldots ,t-1$ and

$$\begin{aligned} \tilde{\sigma _t}=(-1)^t\sigma _1^t-\sigma _{k-1}. \end{aligned}$$

(35)

From (30) and (35) we obtain that in this case

$$\begin{aligned} \varDelta _{k,t}=(-1)^{t+1}\sigma _1^t\sigma _{k-1}^{t-1}+\sigma _{k-1}^t, \end{aligned}$$

thus $\dfrac{\tilde{\sigma _t}}{\varDelta _{k,t}}=\dfrac{-1}{\sigma _{k-1}^{t-1}}$ and we have that

$$\begin{aligned} \tilde{\sigma }_{2t-i}=\tilde{\sigma }_{2t}\overline{\tilde{\sigma _i}}=\dfrac{(-1)^{t}\sigma ^t_k\tilde{\sigma _t}}{\varDelta _{k,t}}\overline{\tilde{\sigma _i}}=(-1)^{t-1+i}\sigma _1^{t-i}\sigma _{k-1} \end{aligned}$$

(36)

for $i=0,\ldots ,t-1$. Finally, we have

$$\begin{aligned} G_{k,t}=\sum _{i=0}^{t-1}\sigma _1^iz^{2t-i}+(\sigma _1^t+(-1)^{t+1}\sigma _{k-1})z^t+(-1)^{t-1}\sum _{i=t+1}^{2t}\sigma _1^i\sigma _{k-1}z^{2t-i} \end{aligned}$$

and performing the product we arrive at the desired result. $\square$

Moreover, we can obtain others group-type t-designs in a similar fashion. Observe that due to $k=t+1$, the obtained t-design X is such that $|X|=3t+1$ then X is the union of, at most, two n-gons. In the case of Proposition 37, we get the union of a 2t-gon and a $t+1$-gon, but we get other combinations if we ask additional algebraic conditions on the $\sigma$’s, as the following examples shows.

Example 38

If we take $k=5$ and $t=4$ and we ask the conditions $\sigma _1=\sigma _4=0$ and $\sigma _5=\sigma _2\sigma _3$ we get

$$\begin{aligned} L(z)G_{5,4}(z)=(z^7+\sigma _2^2\sigma _3)(z^6-\sigma _3^2), \end{aligned}$$

that is, the 4-design obtained is the union of a hexagon and a heptagon.

Example 39

If we take $k=4$ and $t=3$ and we ask the conditions $\sigma _4=\sigma _1\sigma _3$, $\sigma _1^2=\sigma _2$ and $\sigma _3=\sigma _1\sigma _2$ we get

$$\begin{aligned} L(z)G_{4,3}(z)=(z^5+\sigma _1^3\sigma _2)^2, \end{aligned}$$

that is, the 3-design obtained is a pentagon with all points repeated twice.

It would be interesting to have a result describing the general pattern for obtaining group-type t-designs for general k and t.

5 The space of t-designs

To finish this article, we would like to discuss how t-designs in $\mathbb {R}^2$ could be understood as elements of a subset lying in a bigger space. Our results show that when conditions are imposed on the symmetric functions of complex numbers in $S^1$ instead of on the numbers themselves, the existence (or the non-existence) of a t-design and the condition to be of group-type becomes clearer than it would be if we simply analyzed the given finite subset of $S^1$.

It is known (see, for instance [12, Appendix V]) that, if we denote $<a_1,\ldots ,a_k>$ a k-uple in $\mathbb {C}^k$ without regarding the ordering of the elements, the k-th symmetric power of $\mathbb {C}$

$$\begin{aligned} \mathbb {C}_{sym}^k=\{<a_1,\ldots ,a_k>: a_1,\ldots ,a_k\in \mathbb {C}\}, \end{aligned}$$

is in bijective correspondence with $\mathbb {C}^k$ by means of the mapping $\Sigma :\mathbb {C}_{sym}^k\rightarrow \mathbb {C}^k$ given by

$$\begin{aligned} \Sigma (<a_1,\ldots ,a_k>)=(\sigma _1(a_1,\dots ,a_k),\ldots ,\sigma _k(a_1,\ldots ,a_k)). \end{aligned}$$

Moreover, the space $\mathbb {C}_{sym}^k$ can be realized as a analytic variety and the mapping $\Sigma$ is holomorphic. This shows that one can work indistinctly in the set of unordered k-uples of complex numbers or in the $\sigma -$space. Nevertheless, it is not that clear what the rank of the mapping $\Sigma$ is if instead we consider unordered k-uples of elements in $S^1$.

We can obtain a more suitable insight applying Theorem 12, considering each t-design being parametrized by an element of $\mathbb {C}^k_{sym}$ or, equivalently, by the k-uple $(\sigma _1,\ldots ,\sigma _k)$.

Definition 40

Let $\sigma =(\sigma _1,\ldots ,\sigma _{k-1})\in \mathbb {C}^{k-1}$ and $X_k$, $W_k$ as in equations (6), (7), respectively. We say that $\sigma$ is t-compatible if

• 1. $\varDelta _{k,t}(\sigma _1,\ldots ,\sigma _{k-1},(-1)^k)\ne 0$

• 2. $(-1)^i\sigma _i=X_kw$, $w\in W_k$.

We call the set of all t-compatible $\sigma$, the t-compatible space.

Observe that the t-compatible space can be described by means of the set $W_k$.

Proposition 41

For every set $\{a_1,\ldots ,a_k\}$ of points in $S^1$ in t-good position, there exists a $\sigma$ in the t-compatible space.

Proof

Let $a_1,\ldots ,a_k$ in t-good position. By Proposition 35, we can rotate them in order that the resulting set is also in good position and satisfies $\sigma _k=(-1)^k$, where we denote also by $\sigma _i$ the simmetric functions of the rotated points. Thus, the polynomial $z^k-\sigma _1z^{k-1}+\cdots +(-1)^{k-1}\sigma _{k-1}z+1$ is conjugate reciprocal and all of its roots are in the unit cicle, that is, $(-1)^i\sigma _i=X_kw$ for some $w\in W_k$. $\square$

Using Proposition 22 we obtain a nice description of the space of 2-designs with 7 elements.

Corollary 42

For every 2-design with 7 elements such that at least five of them are distinct, we have a $\sigma =(\sigma _1,\bar{\sigma _1})$ in the 2-compatible space.

To give a description of the space of 2-designs with 7 elements we first need to find all $\sigma _1\in \mathbb {C}$ such that $\sigma =(\sigma _1,\bar{\sigma _1})$ is 2-compatible, that is, to find all $(w_1,w_2)\in W_3$, so we consider the polynomial

$$\begin{aligned} z^3-\sigma _1z^2+\bar{\sigma _1}z+1=z^3+(w_1+iw_2)z^2+(w_1-iw_2)z+1. \end{aligned}$$

Since $\sigma _1=-w_1-iw_2$, we consider $\varDelta (-w_1,-w_2)=0$ (cf. eq.(8)) as the boundary of the set of $(w_1,w_2)$ such that the corresponding $\sigma$ are compatible.

• The condition $\varDelta _{3,2}=0$ is $\overline{\sigma _1}^2+\sigma _1=0$, which happens if and only if $\sigma _1\in \{0,1,\frac{1}{2}\pm \frac{\sqrt{3}}{2}\}$, so we exclude these points.

• By Proposition 24, the equation $\sigma _3=\sigma _1\sigma _2$ corresponds to group-type 2-designs, in this case this is equivalent to $|\sigma _1|=1$.

• Finally, in Corollary 16, item 3, we saw that for every $\sigma$ with $|\sigma _1|\le \alpha$, $\alpha \simeq 0.60712$, we obtain a 2-design.

These considerations are reflected in Fig. 3, draw in the w-space. The interior of the curve corresponds to the set of compatible $\sigma$, and the shaded regions correspond to actual 2-designs. Note that there may be other 2-designs besides these ones (see Corollary 16, items 1 and 2).

Fig. 3

The space of 2-designs of 7 elements

We also note that different points in this picture could give raise to the same 2-design. To elucidate this appropriately, we have to decide for what values of $\sigma _1$ the polynomial

$$\begin{aligned}&(z^3-\sigma _1z^2-\overline{\sigma _1}z+1)\left( z^4+\sigma _1z^3+(\sigma _1^2+\overline{\sigma _1})z^2+\frac{\overline{\sigma _1}^2+|\sigma _1|^2\sigma _1}{\overline{\sigma _1}^2+\sigma _1}z+\frac{\sigma _1^2+\overline{\sigma _1}}{\overline{\sigma _1}^2+\sigma _1}\right) \\&\quad =z^7+c_3z^4+c_4z^3+c_0 \end{aligned}$$

is the same one. As $c_4=c_0\overline{c_3}$, it suffices to see for what values of compatible $\sigma _1$

$$\begin{aligned} \begin{array}{cc} \dfrac{\sigma _1^2+\overline{\sigma _1}}{\overline{\sigma _1}^2+\sigma _1}=c_0,&\dfrac{\overline{\sigma _1}^2+|\sigma _1|^2\sigma _1}{\overline{\sigma _1}^2+\sigma _1}-\sigma _1^3-2|\sigma _1|^2+1=c_3 \end{array} \end{aligned}$$

remain equal. For instance, for the values $\frac{1}{3},\frac{-21+\sqrt{513}}{18}\simeq 0.09164$ and $\frac{-21-\sqrt{513}}{18}\simeq -2.425$ of $\sigma _1$ we obtain the same 2-design, that is depicted in Fig. .

This discussion also helps to clarify the question raised in Sect. 3.3: which are all the group-type 2-designs with 7 elements? Solving $c_3c_4=c_0$ we obtain $|\sigma _1|=1$ plus a finite number of points and solving $c_3=c_4=0$ also gives a finite number of points. This illustrates Theorem 3ii) in this case.

Fig. 4

The 2-design whose elements are the zeroes of $z^7+\frac{29}{27}z^4+\frac{29}{27}z^3+1$